Conclusion: For Noetherian $R$, the desired flatness hold only if $R = k$.

Added: A colleague points out that flat local maps of local rings are always faithfully flat,
hence injective. Thus even in the non-Noetherian case, the only way for $k$ to be
flat over $R$ is if $R = k$.

In fact, one can directly extend the above argument to the not-necessarily-Noetherian case, as follows:

Let $I$ be any finitely generated ideal contained in $\mathfrak m$. Since $k$ is
flat, $k\otimes_R I \hookrightarrow k\otimes_R \mathfrak m,$ the target of which vanishes, as noted above.
Thus $k\otimes _R I$ vanishes, and so by Nakayama's lemma, $I = 0$. Since $\mathfrak m$
is the union of such $I$, we see that $\mathfrak m = 0$, and thus $R = k$.